Three charged marbles are glued to a nonconducting surface and are placed in the diagram as shown. The charges of each marble are q1 = 6.20 µC, q2 = 1.53 µC, and q3 = −1.81 µC. Marble q1 is a distance r1 = 3.00 cm to the left of the marble q2, while marble q3 is a distance r3 = 2.00 cm to the right of the marble q2, as shown. Calculate the magnitude of the electric field a distance r' = 1.00 cm to the left of the center marble.

Your answer is 32 hope i  

The answer is frictional force

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j